Mixing
Random Bridge Hand w Cards of exactly two suits
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
Question:
What is the probability that a random bridge hand contains cards of exactly two suits?
My Attempt At A Solution
Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be
$$frac26choose 13-252 choose 13 $$
As user @Lord Shark the Unknown hinted: there are $26choose 13$ ways to choose hands of two suits but one of those suits is only say hearts, and another only spades, so we must compensate for those.
Thank you for any corrections/hints.
probability combinatorics combinations card-games
edited 13 mins ago
N. F. Taussig
40.6k93253
asked 27 mins ago
elcharlosmaster
355
add a comment |Â
up vote
4
down vote
favorite
Question:
What is the probability that a random bridge hand contains cards of exactly two suits?
My Attempt At A Solution
Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be
$$frac26choose 13-252 choose 13 $$
As user @Lord Shark the Unknown hinted: there are $26choose 13$ ways to choose hands of two suits but one of those suits is only say hearts, and another only spades, so we must compensate for those.
Thank you for any corrections/hints.
probability combinatorics combinations card-games
edited 13 mins ago
N. F. Taussig
40.6k93253
asked 27 mins ago
elcharlosmaster
355
3
There are $binom2613$ hands only having spades and hearts. But one of these is all spades, and another all hearts.
– Lord Shark the Unknown
24 mins ago
would it be correct to say $$ frac26 choose 13-252 choose 13$$
– elcharlosmaster
22 mins ago
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Question:
What is the probability that a random bridge hand contains cards of exactly two suits?
My Attempt At A Solution
Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be
$$frac26choose 13-252 choose 13 $$
As user @Lord Shark the Unknown hinted: there are $26choose 13$ ways to choose hands of two suits but one of those suits is only say hearts, and another only spades, so we must compensate for those.
Thank you for any corrections/hints.
probability combinatorics combinations card-games
edited 13 mins ago
N. F. Taussig
40.6k93253
asked 27 mins ago
elcharlosmaster
355
Question:
What is the probability that a random bridge hand contains cards of exactly two suits?
My Attempt At A Solution
Bridge hands consist of $13$ cards, and a suit contains $52$ cards, so the way to pick a random bridge hand would be
$$frac26choose 13-252 choose 13 $$
As user @Lord Shark the Unknown hinted: there are $26choose 13$ ways to choose hands of two suits but one of those suits is only say hearts, and another only spades, so we must compensate for those.
Thank you for any corrections/hints.
probability combinatorics combinations card-games
probability combinatorics combinations card-games
edited 13 mins ago
N. F. Taussig
40.6k93253
asked 27 mins ago
elcharlosmaster
355
edited 13 mins ago
N. F. Taussig
40.6k93253
asked 27 mins ago
elcharlosmaster
355
edited 13 mins ago
N. F. Taussig
40.6k93253
edited 13 mins ago
N. F. Taussig
40.6k93253
edited 13 mins ago
N. F. Taussig
40.6k93253
40.6k93253
asked 27 mins ago
elcharlosmaster
355
asked 27 mins ago
elcharlosmaster
355
asked 27 mins ago
elcharlosmaster
355
355
3
There are $binom2613$ hands only having spades and hearts. But one of these is all spades, and another all hearts.
– Lord Shark the Unknown
24 mins ago
would it be correct to say $$ frac26 choose 13-252 choose 13$$
– elcharlosmaster
22 mins ago
add a comment |Â
3
There are $binom2613$ hands only having spades and hearts. But one of these is all spades, and another all hearts.
– Lord Shark the Unknown
24 mins ago
would it be correct to say $$ frac26 choose 13-252 choose 13$$
– elcharlosmaster
22 mins ago
3
3
There are $binom2613$ hands only having spades and hearts. But one of these is all spades, and another all hearts.
– Lord Shark the Unknown
24 mins ago
There are $binom2613$ hands only having spades and hearts. But one of these is all spades, and another all hearts.
– Lord Shark the Unknown
24 mins ago
would it be correct to say $$ frac26 choose 13-252 choose 13$$
– elcharlosmaster
22 mins ago
would it be correct to say $$ frac26 choose 13-252 choose 13$$
– elcharlosmaster
22 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
7
down vote
accepted
The total number of hands is $$binom5213$$
Now we count the number of hands with cards from exactly two suits.
There are $$binom42$$ ways to choose the two suits.
Given this, there are $$binom2613$$ ways to choose the cards in the hand.
However, we've counted four hands that we shouldn't count: the four hands that have all cards of the same suit (e.g. all $13$ spades).
For each of these $4$ hands, we've counted them $3$ times. For example, $13$ spades and $0$ hearts is counted, but so is $13$ spades and $0$ clubs, and $13$ spades and $0$ diamonds. The number of hands we've overcounted is therefore$$4cdot 3$$
This means that the total number of hands with cards from exactly two suits is
$$binom42binom2613 - 4cdot 3$$
The probability you want is the number of successful possibilities divided by the number of total possibilities:
$$boxed,displaystylefracdisplaystylebinom42displaystylebinom2613 - 4cdot 3displaystylebinom5213,,$$
answered 14 mins ago
Zubin Mukerjee
14.7k32557
4
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
2
Good point, thanks
– Zubin Mukerjee
11 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
accepted
The total number of hands is $$binom5213$$
Now we count the number of hands with cards from exactly two suits.
There are $$binom42$$ ways to choose the two suits.
Given this, there are $$binom2613$$ ways to choose the cards in the hand.
However, we've counted four hands that we shouldn't count: the four hands that have all cards of the same suit (e.g. all $13$ spades).
For each of these $4$ hands, we've counted them $3$ times. For example, $13$ spades and $0$ hearts is counted, but so is $13$ spades and $0$ clubs, and $13$ spades and $0$ diamonds. The number of hands we've overcounted is therefore$$4cdot 3$$
This means that the total number of hands with cards from exactly two suits is
$$binom42binom2613 - 4cdot 3$$
The probability you want is the number of successful possibilities divided by the number of total possibilities:
$$boxed,displaystylefracdisplaystylebinom42displaystylebinom2613 - 4cdot 3displaystylebinom5213,,$$
answered 14 mins ago
Zubin Mukerjee
14.7k32557
4
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
2
Good point, thanks
– Zubin Mukerjee
11 mins ago
add a comment |Â
up vote
7
down vote
accepted
The total number of hands is $$binom5213$$
Now we count the number of hands with cards from exactly two suits.
There are $$binom42$$ ways to choose the two suits.
Given this, there are $$binom2613$$ ways to choose the cards in the hand.
However, we've counted four hands that we shouldn't count: the four hands that have all cards of the same suit (e.g. all $13$ spades).
For each of these $4$ hands, we've counted them $3$ times. For example, $13$ spades and $0$ hearts is counted, but so is $13$ spades and $0$ clubs, and $13$ spades and $0$ diamonds. The number of hands we've overcounted is therefore$$4cdot 3$$
This means that the total number of hands with cards from exactly two suits is
$$binom42binom2613 - 4cdot 3$$
The probability you want is the number of successful possibilities divided by the number of total possibilities:
$$boxed,displaystylefracdisplaystylebinom42displaystylebinom2613 - 4cdot 3displaystylebinom5213,,$$
answered 14 mins ago
Zubin Mukerjee
14.7k32557
4
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
2
Good point, thanks
– Zubin Mukerjee
11 mins ago
add a comment |Â
up vote
7
down vote
accepted
up vote
7
down vote
accepted
The total number of hands is $$binom5213$$
Now we count the number of hands with cards from exactly two suits.
There are $$binom42$$ ways to choose the two suits.
Given this, there are $$binom2613$$ ways to choose the cards in the hand.
However, we've counted four hands that we shouldn't count: the four hands that have all cards of the same suit (e.g. all $13$ spades).
For each of these $4$ hands, we've counted them $3$ times. For example, $13$ spades and $0$ hearts is counted, but so is $13$ spades and $0$ clubs, and $13$ spades and $0$ diamonds. The number of hands we've overcounted is therefore$$4cdot 3$$
This means that the total number of hands with cards from exactly two suits is
$$binom42binom2613 - 4cdot 3$$
The probability you want is the number of successful possibilities divided by the number of total possibilities:
$$boxed,displaystylefracdisplaystylebinom42displaystylebinom2613 - 4cdot 3displaystylebinom5213,,$$
answered 14 mins ago
Zubin Mukerjee
14.7k32557
The total number of hands is $$binom5213$$
Now we count the number of hands with cards from exactly two suits.
There are $$binom42$$ ways to choose the two suits.
Given this, there are $$binom2613$$ ways to choose the cards in the hand.
However, we've counted four hands that we shouldn't count: the four hands that have all cards of the same suit (e.g. all $13$ spades).
For each of these $4$ hands, we've counted them $3$ times. For example, $13$ spades and $0$ hearts is counted, but so is $13$ spades and $0$ clubs, and $13$ spades and $0$ diamonds. The number of hands we've overcounted is therefore$$4cdot 3$$
This means that the total number of hands with cards from exactly two suits is
$$binom42binom2613 - 4cdot 3$$
The probability you want is the number of successful possibilities divided by the number of total possibilities:
$$boxed,displaystylefracdisplaystylebinom42displaystylebinom2613 - 4cdot 3displaystylebinom5213,,$$
answered 14 mins ago
Zubin Mukerjee
14.7k32557
answered 14 mins ago
Zubin Mukerjee
14.7k32557
answered 14 mins ago
Zubin Mukerjee
14.7k32557
answered 14 mins ago
Zubin Mukerjee
14.7k32557
14.7k32557
4
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
2
Good point, thanks
– Zubin Mukerjee
11 mins ago
add a comment |Â
4
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
2
Good point, thanks
– Zubin Mukerjee
11 mins ago
4
4
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
+1 for this. But slightly simpler is to say that for each choice of two suits, there are $binom2613-2$ ways to choose the cards. This gives you the same answer.
– TonyK
12 mins ago
2
2
Good point, thanks
– Zubin Mukerjee
11 mins ago
Good point, thanks
– Zubin Mukerjee
11 mins ago
add a comment |Â
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3
There are $binom2613$ hands only having spades and hearts. But one of these is all spades, and another all hearts.
– Lord Shark the Unknown
24 mins ago
would it be correct to say $$ frac26 choose 13-252 choose 13$$
– elcharlosmaster
22 mins ago